inequality problem

**sarahh** · January 2nd 2010, 06:35 PM

For some dang reason I can't figure out this inequality problem..it's just not working! The solution is $c < -2$ but I keep getting $c > -2$

Problem: $|\frac{4+c} {c}|<1$
(absolute value of the fraction)

**tonio** · January 2nd 2010, 06:56 PM

Originally Posted by sarahh

For some dang reason I can't figure out this inequality problem..it's just not working! The solution is $c < -2$ but I keep getting $c > -2$

Problem: $|\frac{4+c} {c}|<1$
(absolute value of the fraction)

$|\frac{4+c} {c}|<1\Longleftrightarrow -1<\frac{4+c}{c}< 1$ , so the next inequalities have both to be true:

1) $-1< \frac{4+c}{c}\Longrightarrow 2\,\frac{2+c}{c}> 0\Longrightarrow c< -2\,\,\,or\,\,\,c> 0$

2) $\frac{4+c}{c}< 1\Longrightarrow \frac{4}{c}< 0\Longrightarrow c< 0$

From (1)(2) we get $c< -2$

Tonio

**sarahh** · January 2nd 2010, 10:42 PM

Hi Tonio--I still don't understand how you got line (1) and wouldn't line (2) produce $4 < 0$ and not $c < 0$ the former being nonsensical since the variable is gone when multiplied by 0.

Here's why I come out with $c > -2$ :

$|\frac{4+c} {c}|<1$
$\frac{4}{c} + 1>-1$
$\frac{4}{c}>-2$
$(c)\frac{4}{c}>-2c => 4>-2c$
Flip signs since dividing by a negative! $-2<c => c>-2$

**11rdc11** · January 2nd 2010, 11:06 PM

1)
$-1 < \frac{4+c}{c}$

$0 < \frac{4+c}{c} +1$

$0< \frac{4+2c}{c}$

now set your denominator and numerator equal to 0

and you get c equals -2 and 0

Now test a value to the left of -2 on the number line, test a value between -2 and 0, and then test a value to the right of 0 into the inequality

$0< \frac{4+2(-3)}{-3}= \frac{2}{3}$

which makes the inequality true

$0< \frac{4+2(-1)}{-1} = -2$

which makes the inequality false

$0< \frac{4+2(1)}{1)} = 6$

Which makes the inequality true

so $c < -2$ or $c>0$

2)

$1> \frac{4+c}{c}$

$0> \frac{4+c}{c} -1$

$0> \frac{4+c}{c} -\frac{c}{c}$

$0> \frac{4}{c}$

set your numerator and denominator equal to 0 again and you get c = 0

Now test a value to the left of o and to the right of zero

$0> \frac{4}{-1}= -4$

which is true

$0> \frac{4}{1}= 4$

which is false.

so

c<0

Now putting all this you notice the inequality of

$c<0$ and $c<-2$ overlap so

$c<-2$

is your answer.

**11rdc11** · January 2nd 2010, 11:14 PM

Originally Posted by sarahh

Hi Tonio--I still don't understand how you got line (1) and wouldn't line (2) produce $4 < 0$ and not $c < 0$ the former being nonsensical since the variable is gone when multiplied by 0.

Here's why I come out with $c > -2$ :

$|\frac{4+c} {c}|<1$
$\frac{4}{c} + 1>-1$
$\frac{4}{c}>-2$
$(c)\frac{4}{c}>-2c => 4>-2c$
Flip signs since dividing by a negative! $-2<c => c>-2$

The problem with trying to work it out this way is when you are multiplying by c to both sides of the equation cancel out the c in the denominator you assumed c is positive. We do not know if c is positive or negative.

**sarahh** · January 2nd 2010, 11:15 PM

Wow 11, I never saw a problem like this in any math book I've read lol. Very different approach to finding the solution, but I get it! Thanks!

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